Transistors.


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Bipolar Transistors. Two PN intersections joined togetherTwo sorts accessible
Transcripts
Slide 1

Transistors Transfer Resistor Chapter 9

Slide 2

Bipolar Transistors Collector Base Emitter Two PN intersections combined Two sorts accessible – NPN and PNP The areas (start to finish) are known as the authority (C), the base (B), and the emitter (E)

Slide 3

Operation Begin by turn around biasing the CB intersection Here we are demonstrating a NPN transistor for instance Now we apply a little forward predisposition on the emitter-base intersection Electrons are pushed into the base, which then rapidly stream to the gatherer The outcome is a vast emitter-authority electron current (customary current is C-E) which is kept up by a little E-B voltage Some of the electrons pushed into the base by the forward inclination E-B voltage wind up draining gaps in that intersection This would in the end annihilate the intersection in the event that we didn\'t renew the gaps The electrons that may do this are drawn off as a base current

Slide 4

Currents

Slide 5

Conventional View

Slide 6

Origin of the names the Emitter "transmits" the electrons which go through the gadget the Collector "gathers" them again once they\'ve gone through the Base ...and the Base ?...

Slide 7

Original Manufacture

Slide 8

Base Thickness The thickness of the unmodified Base district must be perfect. Too thin, and the Base would basically vanish. The Emitter and Collector would then frame a constant bit of semiconductor, so present would stream between them whatever the base potential. Too thick, and electrons entering the Base from the Emitter wouldn\'t see the Collector as it would be too far away. So then, the current would all be between the Emitter and the Base, and there\'d be no Emitter-Collector current.

Slide 9

Amplification Properties The C-B voltage intersection works close breakdown. This guarantees a little E-B voltage causes torrential slide Large current through the gadget

Slide 10

Common Base NPN

Slide 11

Common Emitter NPN

Slide 12

Common Collector NPN How can I C shift with V CE for different I B ? Take note of that both dc sources are variable Set V BB to build up a specific I B

Slide 13

Collector Characteristic Curve If V CC = 0, then I C = 0 and V CE = 0 As V CC ↑ both V CE and I C ↑ When V CE  0.7 V, base-authority gets to be switch one-sided and I C achieves full esteem (I C = b I B ) I C ~ steady as V CE ↑. There is a slight increment of I C because of the broadening of the consumption zone (BC) giving less openings for recombinations with e ¯ in base. Since I C = b I B , distinctive base streams create diverse I C levels.

Slide 14

NPN Characteristic Curves

Slide 15

PNP Characteristic Curves

Slide 16

Load Line Slope of the heap line is 1/R L For a steady load, venturing I B gives distinctive streams (I C ) anticipated by where the heap line crosses the trademark bend. I C = b I B works inasmuch as the heap line meets on the level district of the bend.

Slide 17

Cut-off Saturation and Cut-off Note that the heap line converges the 75 mA bend underneath the level locale. This is immersion and I C = b I B doesn\'t work in this locale.

Slide 18

Example We alter the base current to 200 m An and take note of that this transistor has a b = 100 Then I C = b I B = 100(200 X 10 - 6 A) = 20 mA Notice that we can utilize Kirchhoff\'s voltage law around the right half of the circuit V CE = V CC – I C R C = 10 V – (20 mA)(220 W ) = 10 V – 4.4 V = 5.6 V

Slide 19

Example Now change I B to 300 m A Now we get I C = 30 mA And V CE = 10 V – (30 mA)(220 W ) = 3.4 V Finally, conform I B = 400 m An I B = 40 mA and V CE = 1.2 V

Slide 20

Plot the heap line

Slide 21

Gain as a component of I C As temperature builds, the pick up increments for every single current esteem.

Slide 22

Operating Limits There will be a breaking point on the scattered power P D(max) = V CE I C V CE and I C were the parameters plotted on the trademark bend. In the event that there is a voltage restrain (V CE(max) ), then you can register the I C that outcomes If there is a present utmost (I C(max) ), then you can process the V CE that outcomes

Slide 23

Example Assume P D(max) = 0.5 W V CE(max) = 20 V I C(max) = 50 mA

Slide 24

Operating Range Operating Range

Slide 25

Voltage Amplifiers Common Base PNP Now we have included an air conditioner source The biasing of the intersections are: BE is forward one-sided by V BB - along these lines a little resistance BC is turn around one-sided by V CC – and a vast resistance Since I B is little, I C  I E

Slide 26

Equivalent air conditioning Circuit r E = inner air conditioning emitter resistance I E = V in/r E (Ohm\'s Law) V out = I C R C  I E R C Recall the name – exchange resistor

Slide 27

Current Gains Common Base a = I C/I E < 1 Common Emitter b = I C/I B

Slide 28

Example If b = 50, then a = 50/51 = 0.98 Recall a < 1 Rearranging, b = a + abdominal muscle b (1-a ) = a b = a/(1-a )

Slide 29

Transistors as Switches

Slide 30

The working focuses We can control the base current utilizing V BB (we don\'t really utilize a physical switch). The circuit then goes about as a rapid switch.

Slide 31

Details In Cut-off All streams are zero and V CE = V CC In Saturation I B sufficiently enormous to create I C(sat)  b I B Using Kirchhoff\'s Voltage Law through the ground circle V CC = V CE(sat) + I C(sat) R C yet V CE(sat) is little (couple of tenths), so I C(sat) V CC/R C

Slide 32

Example What is V CE when V in = 0 V? Ans. V CE = V CC = 10 V b) What least estimation of I B is required to immerse the transistor if b = 200? Take V CE(sat) = 0 V I C(sat)  V CC/R C = 10 V/1000 W = 10 mA Then, I B = I C(sat)/b = 10 mA/200 = 0.05mA

Slide 33

LED Example If a square wave is contribution for V BB , then the LED will be on when the info is high, and off when the information is low.

Slide 34

Transistors with air conditioning Input Assume that b is to such an extent that I C differs somewhere around 20 and 40 mA. The transistor is always showing signs of change bends along the heap line.

Slide 35

Pt. A compares to the positive pinnacle. Pt. B compares to the negative pinnacle. This diagram indicates perfect operation.

Slide 36

Driven to immersion Driven into Cutoff Distortion The area of the point Q (size of the dc source on information) may bring about a working point to lie outside of the dynamic range.

Slide 37

Base Biasing It is generally not important to give two sources to biasing the transistor. The red bolts take after the base-emitter part of the circuit, which contains the resistor R B . The voltage drop crosswise over R B is V CC – V BE (Kirchhoff\'s Voltage Law). The base current is then… and I C = b I B

Slide 38

Base Biasing Use Kirchhoff\'s Voltage Law on the dark arrowed circle of the circuit V CC = I C R C + V CE So, V CE = V CC – I C R C V CE = V CC – b I B R C Disadvantge b happens in the condition for both V CE and I C But b changes – hence do V CE and I C This moves the Q-point ( b - dpendent)

Slide 39

Example Let R C = 560 W @ 25 °C b = 100 R B = 100 k W @ 75 °C b = 150 V CC = +12 V @ 75 °C I B is a similar I C = 16.95 mA V CE = 2.51 V I C increments by half V CE diminishes by 56%

Slide 40

Transistor Amplifiers Amplification The way toward expanding the quality of a flag. The consequence of controlling a generally extensive amount of current (yield) with a little amount of current (info). Speaker Device use to expand the present, voltage, or force of the information motion without obviously changing the basic quality.

Slide 41

Class An Entire info waveform is steadfastly replicated. Transistor invests its whole energy in the dynamic mode Never comes to either cutoff or immersion. Drive the transistor precisely somewhere between cutoff and immersion. Transistor is dependably on – continually dispersing power – can be very wasteful

Slide 42

Class A

Slide 43

Class B No DC predisposition voltage The transistor invests a large portion of its energy in dynamic mode and the other half in cutoff

Slide 44

Push-pull Pair Transistor Q1 "pushes" (drives the yield voltage in a positive heading concerning ground), while transistor Q2 "pulls" the yield voltage (in a negative bearing, toward 0 volts as for ground). Separately, each of these transistors is working in class B mode, dynamic just for one-portion of the info waveform cycle. Together, notwithstanding, they work as a group to deliver a yield waveform indistinguishable fit as a fiddle to the info waveform.

Slide 45

Class AB Between Class A (100% operation) and Class B (half operation).

Slide 46

Class C I C streams for not as much as half then cycle. Normally get more pick up in Class B and C, however more contortion

Slide 47

Common Emitter Transistor Amplifier Notice that V BB forward inclinations the emitter-base intersection and dc current courses through the circuit at all times The class of the enhancer is controlled by V BB as for the info flag. Flag that adds to V BB makes transistor current increment Signal that subtracts from V BB makes transistor current reduction

Slide 48

Details At positive pinnacle of information, V BB is adding to the information Resistance in the transistor is diminished Current in the circuit expands Larger current means more voltage drop crosswise over R C (V RC = IR C ) Larger voltage drop crosswise over R C leaves less voltage to be dropped over the transistor We take the yield V CE – as information builds, V CE diminishes.

Slide 49

More points of interest As the information goes to the negative pinnacle Transistor resistance builds Less present streams Less voltage is dropped crosswise over R C More voltage can be dropped crosswise over C-E The outcome is a stage inversion Feature of the normal emitter intensifier The nearer V BB is to V CC , the bigger the transistor current.

Slide 50

PNP

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